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Question: The centre of a circle passing through the point (0,0), (1,0) and touching the circle \(x^2 + y^2 = ...

The centre of a circle passing through the point (0,0), (1,0) and touching the circle x2+y2=9x^2 + y^2 = 9 is?
A. (32,12)(\dfrac{3}{2}, \dfrac{1}{2})
B. (12,32)(\dfrac{1}{2}, \dfrac{3}{2})
C. (12,12)(\dfrac{1}{2}, \dfrac{1}{2})
D. None of these

Explanation

Solution

Here we will use the basic equation of circle for finding the centre and put the given values in the equation.

Complete step by step solution:
Given: The circle passes through (0,0), (1,0) and touches the circle x2+y2=9x^2 + y^2 = 9.

Let the centre of the circle C be (a, b) and radius r
Circle x2+y2=9{x^2} + {\rm{ }}{y^2} = {\rm{ }}9 has centre (0,0)\left( {0,0} \right){\rm{ }} and radius is 33.
Therefore, the basic equation of a circle C having centre (a, b) and radius rr is (x−a)2+(y−b)2=r2{\left( {x - a} \right)^2} + {\rm{ }}{\left( {y - b} \right)^2} = {\rm{ }}{r^2}
According to question the circle passes through (0,0)
Putting (0,0) in the equation
(0−a)2+(0−b)2=r2{\left( {0 - a} \right)^2} + {\rm{ }}{\left( {0 - b} \right)^2} = {\rm{ }}{r^2}
a2+b2=r2{a^2} + {\rm{ }}{b^2} = {\rm{ }}{r^2} ………. (i)
Also, given the circle passes through (1,0)
Putting in the equation
(1−a)2+(0−b)2=r2{\left( {1 - a} \right)^2} + {\rm{ }}{\left( {0 - b} \right)^2} = {\rm{ }}{r^2}
(1−a)2+(b)2=r2{\left( {1 - a} \right)^2} + {\rm{ }}{\left( b \right)^2} = {\rm{ }}{r^2} ………. (ii)
Subtracting equation (ii) from equation (i)
⇒(1−a)2+(b)2−(a2+b2)=r2−r2\Rightarrow {\left( {1 - a} \right)^2} + {\rm{ }}{\left( b \right)^2}-{\rm{ }}\left( {{a^2} + {\rm{ }}{b^2}} \right){\rm{ }} = {\rm{ }}{r^2} - {\rm{ }}{r^2}

⇒(1−a)2−a2=0\Rightarrow (1-a)^2 - a^2 =0

⇒(1−a)2=a2\Rightarrow (1-a)^2 =a^2

⇒1+a2−2a=a2\Rightarrow 1+a^2-2a=a^2

⇒1−2a=0\Rightarrow 1-2a=0

⇒2a=1\Rightarrow 2a=1

⇒a=12\Rightarrow a= \dfrac{1}{2}

Since it is given that circle C is touching the circle x2+y2=9{x^2} + {\rm{ }}{y^2} = {\rm{ }}9 and given it passes through (0,0), (1,0) this means that a=12a = \dfrac{1}{2} is inside the circle.
This means difference between the centers of the two circles is 3−r3-r
Let D be the difference between the centers of the two circles
D=3−rD{\rm{ }} = {\rm{ }}3 - r
⇒3−r=((a−0)2+(b−0)2)\Rightarrow 3 - r = \sqrt {({{(a - 0)}^2} + {{(b - 0)}^2})}
Squaring both sides
(3−r)2=a2+b2{\left( {3 - r} \right)^2} = {\rm{ }}{a^2} + {\rm{ }}{b^2}
⇒9+r2−6r=a2+b2\Rightarrow 9{\rm{ }} + {\rm{ }}{r^2} - {\rm{ }}6r{\rm{ }} = {\rm{ }}{a^2} + {\rm{ }}{b^2}
Using equation (i)
9+r2−6r=r2    9{\rm{ }} + {\rm{ }}{r^2} - {\rm{ }}6r{\rm{ }} = {\rm{ }}{r^2}\;\;
⇒6r=9\Rightarrow 6r{\rm{ }} = {\rm{ }}9

⇒r=32\Rightarrow r=\dfrac{3}{2}

Putting the values of a and r in (i)

122+b2=322{\dfrac{1}{2}}^2+b^2={\dfrac{3}{2}}^2
⇒b=2\Rightarrow b=2

Therefore, the centre of the circle is (12,2)(\dfrac{1}{2}, 2). So, Option (D) is correct.

Note: In such type of questions we firstly find the centre and radius of the circle and make the equations according to the problem statement.